To find the length of the curve R(t) = (2t + 9, ½ t² − 2, 1/2 √3) where 0 ≤ t ≤ 2, we first need to know that the length of a curve is calculated by finding the integral of the magnitude of the derivative of the function of the curve with respect to the parameter, from the start to the end of the parameter's limits.
In simpler terms, we will first have to differentiate R(t) into R'(t), find the magnitude of the derivative, then integrate from t = 0 to t = 2.
So, let's start:
Step 1: Differentiate R(t) with respect to t.
R'(t) = d/dt [2t + 9, ½ t² − 2, 1/2 √3]
We are differentiating each component of this vector with respect to t, which gives us:
R'(t) = [2, t, 0]
Step 2: We need to find the magnitude (or the length) of the vector R'(t). This can be found using the formula for the length of a vector, which is the square root of the sum of the squares of its components.
| R'(t) | = √(2² + t² + 0²)
= √(4 + t²)
= 2 √(0.25t² + 1)
Step 3: Finally, we find the length of the curve between t=0 and t=2. This is calculated by integrating | R'(t) | from 0 to 2.
∫ from 0 to 2 √(4 + t²) dt
= ∫ from 0 to 2 2 √(0.25t² + 1) dt
= 4.59 (rounded to 2 decimal places).
So, The length of the curve R(t) = (2t + 9, ½ t² − 2, 1/2 √3) from t=0 to t=2 is approximately 4.59 units.